I am having a hard time understanding the intuition for solving problems asking for the probability of a sequence given a number of stages and a number of elements. Specifically, I struggle with differentiating between stages and elements.
I had a homework problem along with two solutions: a counting method and a binomial method. Intuitively, I understand the binomial method better- I can identify what a success is, how many times I would like it to happen, what a failure is, and how many times I would like it to happen. However, my understanding of using the counting method limits me.
The homework problem was to determine the probability that exactly one customer of six customers chooses to dine on the 1st floor of three different floors. The counting solution was to start with the number of different outcomes.
I am having a hard time grasping the difference between a floor selecting up to six people and a person selecting a floor. Is the difference that whatever is doing the selecting can only actually make one selection? In this case, a floor can have multiple people but a person can only have one floor: therefore a person is choosing a floor and not a floor that is choosing a person. I know that the formula for number of outcomes is r^n, or, number of stages to the power of number of elements. How do I know that stage 1 is not Person 1, and that the number of outcomes is not 6^3?
For similar reasons, I struggle with the binomial solution as well. Does (6c1) refer to 6 people choosing a single floor each? I saw a post that said to think of (6c1) as choosing a customer, not a floor, which I do not understand as there are not 6 floors choosing 1 customer. Also, for the binomial solution, if the 1st floor has 6 choices of customers and is therefore (6c1), why is the number of remaining people choosing one of two remaining floors not (5c2) instead of 2^5? Or, on the same token, why is (6c1) not 1^6 (what I am getting as is the difference between these two form factors).
Any help is appreciated. I have been looking through my book and (felt like) I was understanding the introductory examples, but then can never sort things out when I get to harder problems.
Edit:
From my understanding of the replies, there are 3 stages because that is how many decisions are being made. The 1st floor decides to take on a value, that is how many people will be on that floor. This floor can be seen as the 1st decision or 1st stage (I understand that the 1st floor being the first decision is arbitrary). That is why, using the formula r^n number of outcomes, there are 3^6 different outcomes- 6 choices (elements) for each of 3 stages (trials). The binomial coefficient, (nck), for the binomial formula in this case is (6c1), because we want to know the number of ways to select 1 person of 6 for a trial we are interested in (the trial of the first floor). Using the counting method, the number of ways to select 1 person is (6c1), and the number of ways to distribute 5 people among 2 floors is 2^5, because each person has a choice of two floors.
I thought that I was understanding the difference between stages and elements now, but I had a hard time comparing this problem to another, similar problem to which I have the solution. The other problem asks what the probability is of rolling two 3's of a 3-sided die in 6 rolls. The probability of rolling a three is 1/4, and therefore the binomial formula is (6c2)*(1/4)^2(3/4)^4. Again, this formula makes sense to me because I can identify how many successes I want and how many not successes I want. Because the coefficient is (6c2), I am guessing that 6 is the number of trials (rolls), and 2 is the number of rolls of interest. To compare this to the 1st problem, this would make a roll a trial, and that trial can take on one of three values (choices), similar to how a floor is a trial, and that trial can take on a value of 0-6 people (choices).
However, if these problems were similar in the way I am thinking they are similar, then in the floor problem, floor=stage=trial, and in the die problem, side of the die=stage=trial. But how can a side of the die be a trial? Is the reason I am confusing this because the 6 people in the restaurant problem are not analogous to the 6 rolls in the die problem, and that instead the 6 people is analogous to the 3 sides of the die (as I suspect)? In that case, I am guessing that the total number of outcomes is 6^3 in the die problem, because each of 6 trials (stages) has three choices.